Question

If the domain of the function $f\left(x\right)={\log }_e\left({\log }_{|\cos x|}\right(x^2-7x+26)-\frac{4}{{\log }_2|\cos x|})$ is set $A,$ then $A$ contains the interval(s)

• A. $[\frac{3\pi }{2},5)$
• B. $(2,\pi )$
• C. $(2,5)$
• D. $(\pi ,\frac{3\pi }{2})$

Answer 1

Answer: (B), (D)

$(\pi ,\frac{3\pi }{2})$

$f\left(x\right)={\log }_e\left({\log }_{|\cos x|}\right(x^2-7x+26)-\frac{4}{{\log }_2|\cos x|})$

For domain of $f$,
$|\cos x|\ne 0,1\Rightarrow x\in R-\left\{\frac{n\pi }{2}\right\}$

Now,$f\left(x\right)={\log }_e\left({\log }_{|\cos x|}\right(x^2-7x+26)-4{\log }_{|\cos x|}2)$
$\Rightarrow f\left(x\right)={\log }_e\left({\log }_{|\cos x|}\left(\frac{x^2-7x+26}{16}\right)\right)$
So, $\frac{x^2-7x+26}{16}>0$
$\Rightarrow x^2-7x+26>0\Delta =49-4\times 24<0$
$\therefore x^2-7x+26>0$ which is true for all real values of $x$.

Also, ${\log }_{|\cos x|}\left(\frac{x^2-7x+26}{16}\right)>0$
As the base of the log function is less than $1$, the inequality sign reverses.
$\Rightarrow \frac{x^2-7x+26}{16}<1$
$\Rightarrow x^2-7x+10<0\Rightarrow (x-2)(x-5)<0\Rightarrow x\in (2,5)$

But $x\ne \pi ,\frac{3\pi }{2}$
Hence, the domain of $f$ is $A=(2,\pi )\cup (\pi ,\frac{3\pi }{2})\cup (\frac{3\pi }{2},5)$